How can I now show that $g(t)-g(s) \leq \int_s^t f(x) \ dx$

Question

Let $f,g \in L^1(a,b)$ such that $$-\int_a^b\phi'(x)g(x)\ dx \leq \int_a^b\phi(x)f(x)\ dx $$ for all $\phi \in C_c^{\infty}(a,b)$ with $\phi(x) \geq 0$ for $x \in (a,b)$. How can I now show that $$g(t)-g(s) \leq \int_s^t f(x) \ dx$$ for almost all $s,t \in (a,b)$?

I little help on how I can start would be much appreciated.

Answer 1

It suffices to show that for all $\varphi,\psi\in C_0^\infty(a,b)$ $$ \int_a^b\Big(\int_a^t \big(g(t)-g(s)\big)\varphi(s)\,ds\big)\psi(t)\,dt\le \int_a^b\Big(\int_a^t\big(\int_s^t f(x)\,dx\big)\varphi(s)\,ds\Big)\psi(t)\,dt. $$

Let $j_\varepsilon\in C_0^\infty(\mathbb R)$, such that $j_\varepsilon(x)\ge 0$, for all $x$, $\int j_\varepsilon=1$ and $\,\mathrm{supp}\,j_\varepsilon\subset (-\varepsilon,\varepsilon)$. Then for $a<s<t<b,$ set $$ \zeta_\varepsilon(x;s,t)=\int_a^x\big(j_\varepsilon(\xi-s)-j_\varepsilon(\xi-t)\big)\,d\xi. $$ Clearly, $0\le \zeta_\varepsilon(x;s,t)\le 1$, and $\,\mathrm{supp}\,\zeta_\varepsilon \subset (s-\varepsilon,t+\varepsilon)$, while $$ \zeta_\varepsilon(x;s,t)=1, \quad \text{when} \quad x\in [s+\varepsilon,t-\varepsilon]. $$ Next, observe that if $h\in C(a,b)$, then $$ \lim_{\varepsilon\to 0}\int_a^b \zeta_{\varepsilon}'(x;s,t)h(x)\,dx=h(s)-h(t) $$ Now $$ \int_a^b\Big(\int_a^t \big(g(t)-g(s)\big)\varphi(s)\,ds\big)\psi(t)\,dt\\ = -\lim_{\varepsilon\to 0}\int_a^b\Big(\int_a^t \big( \int_a^b\zeta_{\varepsilon}'(x;s,t)g(x)\,dx\big)\varphi(s)\,ds\big)\psi(t)\,dt\\ = \lim_{\varepsilon\to 0}\int_a^b\Big(\int_a^t \big( \int_a^b\zeta_{\varepsilon}(x;s,t)f(x)\,dx\big)\varphi(s)\,ds\big)\psi(t)\,dt\\ = \int_a^b\Big(\int_a^t\big(\int_s^t f(x)\,dx\big)\varphi(s)\,ds\Big)\psi(t)\,dt. $$

Answer 2

Idea:

1. Integrate by parts in the RHS to get $$\int_a^b\phi'(x)[g(x)-F(x)]\,dx\ge 0,\quad\forall \phi$$ where $F(x)=\int_a^xf(t)\,dt$.
2. Make a smooth approximation of $\delta_s-\delta_t$ and take it as $\phi'$. For the final estimation, use this result.

P.S. It looks like a generalization of du Bois-Reymond lemma where it is equality. I guess much of the proof can be adapted to this case as well.